One non-zero digit, one vowel and one consonant from English alphabet (in capital) are to be used in forming passwords, such that each password has to start with a vowel and end with a consonant. How many such passwords can be generated?

The correct answer is (C) 945.

Explanation:

To find the total number of passwords, we must analyze the choices available for each of the three positions: a vowel, a non-zero digit, and a consonant. First, total non-zero digits from 1 to 9 are 9. Second, total vowels in the English alphabet (A, E, I, O, U) are 5. Third, total consonants in the English alphabet are 21. According to the problem, the password consists of 3 characters: position 1 must be a vowel, position 2 can be any of the remaining items, and position 3 must be a consonant. However, let us examine the exact arrangement. The positions are 3 in total: a vowel, a consonant, and a non-zero digit. The password must start with a vowel and end with a consonant. Thus, position 1 can be filled in 5 ways (vowels). Position 3 can be filled in 21 ways (consonants). Position 2 can be filled by the remaining category, which is the non-zero digit. There are 9 non-zero digits. But wait, any of the three characters (one non-zero digit, one vowel, one consonant) are arranged in 3 positions. The first position is a vowel (5 choices), the last position is a consonant (21 choices). The middle position is a non-zero digit (9 choices). Therefore, the total number of passwords = 5 (for vowel) × 9 (for digit) × 21 (for consonant) = 1050. Thus, option D is correct.